What is the locus of the points of intersection of the lines as shown in the figure?

Question

I don't know if it should matter but the sum of the intercepts the lines make with x and y axis is constant I think. It looks like a hyperbola to me.

Answer 1

Parabola if you are joining equispaced divisions by a straight line.

EDIT 1:

The notches or holes are assumed equi-spaced. Let the segments be $ a-u, a+u $ along y and x-axes.

$$ \frac{x}{a-u} + \frac{x}{a+u}= 1 \tag{1}$$

To find envelope by C-discriminant method to find envelope.

$$ \frac{x}{(a-u)^2} - \frac{x}{(a+u) ^2} = 1 \tag{2} $$

$$ \frac{\sqrt x}{\sqrt y }= \frac{a-u}{a+u} \tag{3} $$

$$ \frac{u}{a} = \frac{{\sqrt y}-{\sqrt x} }{{\sqrt y}+{\sqrt x} }\tag{4} $$

Plug (1) into (4) and simplify to get a parabola envelope tangent to axes:

$$ {\sqrt x}+{\sqrt y}= \sqrt { 2 a}. \tag{5} $$

which is a parabola with axes at $45^0$ to x,y axes and tangent to them.

Answer 2

We are given a family of lines $$\ell_c:\quad{x\over1+c}+{y\over 1-c}=1\qquad(-1<c<1)\ .$$ A given point $(x,y)$ belongs to one of the lines iff the equation $$x(1-c)+y(1+c)=1-c^2\tag{1}$$ has real solutions $c$. The boundary between the covered and the uncovered part of the plane is given by the points $(x,y)$ for which the discriminant of the quadratic equation $(1)$ vanishes. This amounts to the equation $$(y-x)^2-4(x+y-1)=0\ .\tag{2}$$ After introducing new orthonormal coordinates $$u:={1\over\sqrt{2}}(y-x),\qquad v:={1\over\sqrt{2}}(y+x)$$ equation $(2)$ assumes the form $v=p u^2 +q$ with certain constants $p$ and $q$. It follows that the locus in question is an arc of a parabola.